# An odd series: $\sum_{n \geq 0} \cos\left(\frac{\pi}{4}\alpha^n\right)$

Let $$\alpha = 7+4\sqrt{3}$$. Consider the series $$\sum_{n\geq 0}\cos\left(\frac{\pi}{4}\alpha^n\right)$$

I'm having a doubt. On one hand, I've plotted (I think) the evolution of the partial sum $$\left(S_n\right)$$ and I've obtained:

So I've assumed that it was divergent. However, it can be easily shown that $$\frac{1}{2}\left(\alpha^n+\frac{1}{\alpha^n}\right) = 2p_n+1$$ where $$p_n$$ is a sequence positive integers. As a result $$\cos\left(\frac{\pi}{4}\alpha^n\right) = \cos\left(\frac{\pi}{4}\left[4p_n+2-\frac{1}{\alpha^n}\right]\right) = \cos\left(\frac{\pi}{2}+\pi p_n -\frac{\pi}{4\alpha^n}\right)=\left(-1\right)^{p_n}\sin\left(\frac{\pi}{4\alpha^n}\right)$$ Noting $$\alpha \approx 14 > 1$$, I've concluded $$\left|\cos\left(\frac{\pi}{4}\alpha^n\right)\right| \underset{(+\infty)}{\sim}\frac{\pi}{4}\frac{1}{\alpha^n}$$ which is then is the general term for a convergent series.

So does it converge or diverge? Could someone point me in the right direction?

• You almost did it. Note that the absolute value decreases to $0$. Commented Jul 6, 2022 at 21:29
• The word is "series" (that's both singular and plural), not "serie". Commented Jul 6, 2022 at 21:30
• it surely converges since it is absolutely convergent as you've shown Commented Jul 6, 2022 at 21:31
• When computing the terms numerically, it is possible there is some rounding error in computing $\alpha^n$, which can result in the calculated terms not approaching zero. Commented Jul 6, 2022 at 21:31